Question

已知数列an中，a1=3,a2=5,其前n项和Sn满足Sn+Sn-2=2Sn-1+2^n-1(n>=3).

令bn=2^n-1/an*an+1;Tn是数列{bn}的前n项和，证明Tn<1/6
要有详细的解析哦

Answer 1

由题意
Sn-Sn-1=Sn-1-Sn-2+2^(n-1)
∴an=an-1+2^(n-1)

an-an-1=2^(n-1)
an-1-an-2=2^(n-2)
……
a2-a1=2
累加得
an-a1=2+2^2+……+2^(n-1)=2^n-2

∴an=2^n+1

∴bn=2^n-1/(2^n+1)(2^(n+1)+1)
=2^n+1/(2^n+1)-2/(2^(n+1)+1)
<2^n+1/(2^n+1)-1/(2^(n+1)+1)

∴Tn<2(1-2^n)/(1-2)+1/3-1/5+1/5-1/9+……+1/(2^n+1)-1/(2^(n+1)+1)
=2^(n+1)-2+1/3-1/(2^(n+1)+1)
=2^(n+1)-1/(2^(n+1)+1)-5/3

再用不等式放缩一下就行了

Answer 2

Sn+Sn-2=2Sn-1+2^n-1
移项可得

Sn-Sn-1=Sn-1 - Sn-2 +2^(n-1)
即
an=an-1 + 2^(n-1) (n>=3)

所以数列an的通项为
an= a1+ 2^1+2^2+...+2^(n-1)
= a1+ 2^n-2 (n>=3)
此通项可与a1=3,a2=5合并，得到

an=2^n+1
则有
bn=2^n-1/(2^n+1)(2^(n+1)+1)
=0.5*[1/(2^n+1)-1/(2^(n+1)+1)]

所以Tn=0.5*[1/3-1/5+1/5-1/9+...+1/(2^n+1)-1/(2^(n+1)+1)] ---注意中间项都正负抵消

=0.5*[1/3-1/(2^(n+1)+1)]
<0.5*1/3= 1/6

证明完毕
PS：以后记得，下标与指数都要加括号呀